Electromagnetic Waves in Vacuum

Step 1: Write Maxwell's equations in vacuum

$$\nabla \cdot \mathbf{E} = 0, \quad \nabla \cdot \mathbf{B} = 0$$

$$\nabla \times \mathbf{E} = -\frac{\partial \mathbf{B}}{\partial t}, \quad \nabla \times \mathbf{B} = \mu_0\epsilon_0\frac{\partial \mathbf{E}}{\partial t}$$

Step 2: Take the curl of Faraday's law

$$\nabla \times (\nabla \times \mathbf{E}) = \nabla \times \left(-\frac{\partial \mathbf{B}}{\partial t}\right) = -\frac{\partial}{\partial t}(\nabla \times \mathbf{B})$$

Step 3: Apply the vector identity on the left side

$$\nabla \times (\nabla \times \mathbf{E}) = \nabla(\nabla \cdot \mathbf{E}) - \nabla^2\mathbf{E} = -\nabla^2\mathbf{E}$$

since $\nabla \cdot \mathbf{E} = 0$ in vacuum (Gauss's law with no charges).

Step 4: Substitute Ampere-Maxwell law on the right side

$$-\frac{\partial}{\partial t}(\nabla \times \mathbf{B}) = -\frac{\partial}{\partial t}\left(\mu_0\epsilon_0\frac{\partial \mathbf{E}}{\partial t}\right) = -\mu_0\epsilon_0\frac{\partial^2 \mathbf{E}}{\partial t^2}$$

Step 5: Combine to get the wave equation

$$\nabla^2\mathbf{E} = \mu_0\epsilon_0\frac{\partial^2 \mathbf{E}}{\partial t^2}$$

This is the electromagnetic wave equation. Comparing with the standard wave equation $\nabla^2 f = \frac{1}{v^2}\frac{\partial^2 f}{\partial t^2}$, we identify:

$$v = \frac{1}{\sqrt{\mu_0\epsilon_0}} = c = 2.998 \times 10^8 \text{ m/s}$$

Step 6: Identical equation for B

Taking the curl of Ampere-Maxwell law and using the same procedure yields:

$$\nabla^2\mathbf{B} = \mu_0\epsilon_0\frac{\partial^2 \mathbf{B}}{\partial t^2}$$

Both $\mathbf{E}$ and $\mathbf{B}$ satisfy the same wave equation and travel at the same speed $c$.

Step 1: Propose a plane wave ansatz

$$\mathbf{E}(z,t) = E_0\cos(kz - \omega t + \phi)\,\hat{x}$$

where $k$ is the wave number, $\omega$ is the angular frequency, and $\phi$ is a phase constant.

Step 2: Compute the spatial derivatives

$$\nabla^2\mathbf{E} = \frac{\partial^2 E_x}{\partial z^2}\,\hat{x} = -k^2 E_0\cos(kz - \omega t + \phi)\,\hat{x} = -k^2\mathbf{E}$$

Step 3: Compute the time derivatives

$$\frac{\partial^2\mathbf{E}}{\partial t^2} = -\omega^2 E_0\cos(kz - \omega t + \phi)\,\hat{x} = -\omega^2\mathbf{E}$$

Step 4: Substitute into the wave equation

$$-k^2\mathbf{E} = \mu_0\epsilon_0(-\omega^2\mathbf{E}) \implies k^2 = \mu_0\epsilon_0\,\omega^2$$

Step 5: Obtain the dispersion relation

$$\omega = \frac{k}{\sqrt{\mu_0\epsilon_0}} = ck$$

This is a linear dispersion relation, meaning EM waves in vacuum are non-dispersive: all frequencies travel at the same speed $c$.

Step 6: Relate wavelength, frequency, and wave number

$$k = \frac{2\pi}{\lambda}, \quad \omega = 2\pi f, \quad \lambda f = c$$

Step 1: Write E for a plane wave in the z-direction

$$\mathbf{E} = E_0\cos(kz - \omega t)\,\hat{x}$$

Step 2: Apply Faraday's law

$$\nabla \times \mathbf{E} = -\frac{\partial \mathbf{B}}{\partial t}$$

Step 3: Compute the curl of E

$$\nabla \times \mathbf{E} = \frac{\partial E_x}{\partial z}\,\hat{y} = -kE_0\sin(kz - \omega t)\,\hat{y}$$

(Only the $\partial E_x/\partial z$ term survives since $E_x$ depends only on $z$ and $t$.)

Step 4: Integrate to find B

$$-\frac{\partial \mathbf{B}}{\partial t} = -kE_0\sin(kz - \omega t)\,\hat{y}$$

$$\mathbf{B} = -\frac{k}{\omega}E_0\cos(kz - \omega t)\,\hat{y} = \frac{E_0}{c}\cos(kz - \omega t)\,\hat{y}$$

where we used $k/\omega = 1/c$.

Step 5: Identify the key relationships

$$|\mathbf{B}| = \frac{|\mathbf{E}|}{c}, \quad \mathbf{B} = \frac{1}{c}(\hat{k} \times \mathbf{E})$$

$\mathbf{E}$ and $\mathbf{B}$ are perpendicular to each other and to $\hat{k}$, and they oscillate in phase.

Step 6: Verify with Ampere-Maxwell law

$$\nabla \times \mathbf{B} = \frac{\partial B_y}{\partial z}\,(-\hat{x}) = \frac{k}{c}E_0\sin(kz-\omega t)\,(-\hat{x})$$

Wait -- let us be careful with the curl components:

$$(\nabla \times \mathbf{B})_x = -\frac{\partial B_y}{\partial z} = \frac{k}{c}E_0\sin(kz-\omega t)$$

$$\mu_0\epsilon_0\frac{\partial E_x}{\partial t} = \frac{1}{c^2}\omega E_0\sin(kz-\omega t) = \frac{k}{c}E_0\sin(kz-\omega t) \;\checkmark$$

Step 1: Write the electromagnetic energy density

$$u = \frac{1}{2}\left(\epsilon_0 E^2 + \frac{B^2}{\mu_0}\right)$$

Step 2: Substitute the plane wave fields

For $E = E_0\cos(kz-\omega t)$ and $B = E_0/c\cdot\cos(kz-\omega t)$:

$$u = \frac{1}{2}\epsilon_0 E_0^2\cos^2(kz-\omega t) + \frac{E_0^2}{2\mu_0 c^2}\cos^2(kz-\omega t)$$

Step 3: Show electric and magnetic energy densities are equal

Since $c^2 = 1/(\mu_0\epsilon_0)$, we have $1/(\mu_0 c^2) = \epsilon_0$, so:

$$u_E = \frac{1}{2}\epsilon_0 E_0^2\cos^2(kz-\omega t) = u_B$$

$$u = \epsilon_0 E_0^2\cos^2(kz-\omega t)$$

Step 4: Compute the Poynting vector

$$\mathbf{S} = \frac{1}{\mu_0}\mathbf{E} \times \mathbf{B} = \frac{1}{\mu_0}E_0\cos(kz-\omega t)\,\hat{x} \times \frac{E_0}{c}\cos(kz-\omega t)\,\hat{y}$$

$$\mathbf{S} = \frac{E_0^2}{\mu_0 c}\cos^2(kz - \omega t)\,\hat{z}$$

Step 5: Time-average the Poynting vector

Using $\langle\cos^2\rangle = 1/2$:

$$\langle\mathbf{S}\rangle = \frac{E_0^2}{2\mu_0 c}\,\hat{z} = \frac{1}{2}\epsilon_0 c E_0^2\,\hat{z}$$

The intensity (time-averaged power per unit area) is $I = |\langle\mathbf{S}\rangle| = E_0^2/(2\mu_0 c)$.

Step 6: Electromagnetic momentum density

$$\mathbf{g} = \mu_0\epsilon_0\mathbf{S} = \frac{\mathbf{S}}{c^2}$$

The momentum carried per unit area per unit time is $\mathbf{S}/c$.

Step 7: Derive radiation pressure for absorption

When a wave is fully absorbed, all its momentum $\Delta p = S\Delta t\,A/c^2 \cdot c = S\,A\,\Delta t/c$ is transferred:

$$P_{\rm rad} = \frac{F}{A} = \frac{\Delta p}{A\,\Delta t} = \frac{\langle S \rangle}{c} = \frac{I}{c}$$

For perfect reflection, the momentum reversal doubles the force: $P_{\rm rad} = 2I/c$.

Step 1: Write the most general transverse E field

$$\tilde{\mathbf{E}} = \left(\tilde{E}_{0x}\,\hat{x} + \tilde{E}_{0y}\,\hat{y}\right)e^{i(kz - \omega t)}$$

where $\tilde{E}_{0x} = E_{0x}e^{i\phi_x}$ and $\tilde{E}_{0y} = E_{0y}e^{i\phi_y}$ are complex amplitudes.

Step 2: Define the Jones vector

The polarization state is fully characterized by the Jones vector (ignoring the common propagation factor):

$$\mathbf{J} = \begin{pmatrix} \tilde{E}_{0x} \\ \tilde{E}_{0y} \end{pmatrix} = \begin{pmatrix} E_{0x}e^{i\phi_x} \\ E_{0y}e^{i\phi_y} \end{pmatrix}$$

Step 3: Linear polarization

When $\phi_x = \phi_y$ (or differ by $\pi$), the field oscillates along a fixed direction. Setting $\phi_x = 0$:

$$\mathbf{J}_{\rm linear} = \begin{pmatrix} \cos\alpha \\ \sin\alpha \end{pmatrix}$$

where $\alpha$ is the polarization angle. Examples: $\hat{x}$-polarized $\to \begin{pmatrix}1\\0\end{pmatrix}$, $\hat{y}$-polarized $\to \begin{pmatrix}0\\1\end{pmatrix}$.

Step 4: Circular polarization

Set $E_{0x} = E_{0y} = E_0/\sqrt{2}$ and $\delta = \phi_y - \phi_x = \pm\pi/2$:

$$\mathbf{J}_{\rm RCP} = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ -i \end{pmatrix}, \qquad \mathbf{J}_{\rm LCP} = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ +i \end{pmatrix}$$

The real fields are $\mathbf{E} = E_0[\cos(kz-\omega t)\,\hat{x} \mp \sin(kz-\omega t)\,\hat{y}]$. The tip of $\mathbf{E}$ traces a circle.

Step 5: Elliptical polarization (general case)

For arbitrary $E_{0x}$, $E_{0y}$, and relative phase $\delta = \phi_y - \phi_x$:

$$\mathbf{J}_{\rm elliptical} = \begin{pmatrix} E_{0x} \\ E_{0y}e^{i\delta} \end{pmatrix}$$

The tip of $\mathbf{E}$ traces an ellipse. Eliminate $t$ from the two components to get the polarization ellipse equation:

$$\frac{E_x^2}{E_{0x}^2} + \frac{E_y^2}{E_{0y}^2} - \frac{2E_x E_y}{E_{0x}E_{0y}}\cos\delta = \sin^2\delta$$

Step 6: Special cases recovered

$\delta = 0$ or $\pi$: ellipse degenerates to a line (linear polarization).

$\delta = \pm\pi/2$ and $E_{0x} = E_{0y}$: ellipse becomes a circle (circular polarization).

Any polarization can be decomposed: $\mathbf{J} = c_+\mathbf{J}_{\rm RCP} + c_-\mathbf{J}_{\rm LCP}$.